Problem: The lifespans of sloths in a particular zoo are normally distributed. The average sloth lives $18.2$ years; the standard deviation is $3.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a sloth living longer than $7.1$ years.
Explanation: $18.2$ $14.5$ $21.9$ $10.8$ $25.6$ $7.1$ $29.3$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $18.2$ years. We know the standard deviation is $3.7$ years, so one standard deviation below the mean is $14.5$ years and one standard deviation above the mean is $21.9$ years. Two standard deviations below the mean is $10.8$ years and two standard deviations above the mean is $25.6$ years. Three standard deviations below the mean is $7.1$ years and three standard deviations above the mean is $29.3$ years. We are interested in the probability of a sloth living longer than $7.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the sloths will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the sloths will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $7.1$ years and the other half $({0.15\%})$ will live longer than $29.3$ years. The probability of a particular sloth living longer than $7.1$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.